Answer:
69.01 m
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you ...
Tan = Opposite/Adjacent
The tangent function is useful for problems like this. Let the height of the spire be represented by h. The distance (d) across the plaza from the first surveyor satisfies the relation ...
tan(50°) = (h -1.65)/d
Rearranging to solve for d, we have ...
d = (h -1.65)/tan(50°)
The distance across the plaza from the second surveyor satisfies the relation ...
tan(30°) = (101.65 -h)/d
Rearranging this, we have ...
d = (101.65 -h)/tan(30°)
Equating these expressions for d, we can solve for h.
(h -1.65)/tan(50°) = (101.65 -h)/tan(30°)
h(1/tan(50°) +1/tan(30°)) = 101.65/tan(30°) +1.65/tan(50°)
We can divide by the coefficient of h and simplify to get ...
h = (101.65·tan(50°) +1.65·tan(30°))/(tan(30°) +tan(50°))
h ≈ 69.0148 ≈ 69.01 . . . . meters
The tip of the spire is 69.01 m above the plaza.
Answer:
Yes, the sum of any two lengths is greater than the third length
Step-by-step explanation:
The sides of the rectangle are aligned with the coordinate axes, so the dimensions of the rectangle are easily determined to be 4 -(-7) = 11 units by 1 -(-1) = 2 units. As with any rectangle, the area is the product of length and width.
The area is 11*2 = 22 square units.
Answer:
yes at negative x-coordnintn
Step-by-step explanation: